Resistance of a 0.1 M KCl solution in a conductance cell is 300 ohm and specific conductance of `"0.1 M KCl"` is `.133xx10^(-2)" ohm"^(-1)"cm"^(-1)`. The resistance of 0.1 M NaCl solution in the same cell is 400 ohm. The equivalent conductance of the 0.1 M NaCl solution `("in ohm"^(-1)"cm"^(2)"/gmeq.")` is

Answer

To find the equivalent conductance of a 0.1 M NaCl solution in the given conductance cell, we will follow these steps:

### Step 1: Understand the given data
- **Resistance of 0.1 M KCl solution (R₁)** = 300 ohm
- **Specific conductance of 0.1 M KCl (κ₁)** = 1.33 × 10^(-2) ohm^(-1) cm^(-1)
- **Resistance of 0.1 M NaCl solution (R₂)** = 400 ohm

### Step 2: Calculate the cell constant (K)
The cell constant (K) can be calculated using the formula:
\[ K = \kappa \times R \]
Where:
- κ is the specific conductance
- R is the resistance

For KCl:
\[ K = \kappa_1 \times R_1 \]
\[ K = (1.33 \times 10^{-2} \, \text{ohm}^{-1} \text{cm}^{-1}) \times 300 \, \text{ohm} \]
\[ K = 1.33 \times 300 \times 10^{-2} \]
\[ K = 399 \, \text{cm}^{-1} \]

### Step 3: Calculate the specific conductance (κ₂) for NaCl
Using the cell constant (K) and the resistance of NaCl (R₂), we can find the specific conductance of the NaCl solution:
\[ \kappa_2 = \frac{K}{R_2} \]
\[ \kappa_2 = \frac{399 \, \text{cm}^{-1}}{400 \, \text{ohm}} \]
\[ \kappa_2 = 0.9975 \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 4: Calculate the equivalent conductance (Λ) of NaCl
The equivalent conductance (Λ) can be calculated using the formula:
\[ \Lambda = \frac{\kappa \times 1000}{n} \]
Where:
- n is the normality of the solution. For 0.1 M NaCl, since NaCl dissociates into one Na⁺ and one Cl⁻, the n factor is 1. Therefore, normality (n) = 0.1.

Substituting the values:
\[ \Lambda = \frac{0.9975 \times 1000}{0.1} \]
\[ \Lambda = \frac{997.5}{0.1} \]
\[ \Lambda = 9975 \, \text{ohm}^{-1} \text{cm}^2/\text{gmeq} \]

### Final Result
The equivalent conductance of the 0.1 M NaCl solution is:
\[ \Lambda = 9975 \, \text{ohm}^{-1} \text{cm}^2/\text{gmeq} \]

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